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12:27 AM
0
Q: Are questions about the field of physics off topic?

M BarbosaI often have questions about the field of physics itself (as opposed to physics questions). For example questions about the current state of a sub-field of physics, or questions about the implications of experiment results. The latter is especially important to me because my knowledge of physics ...

discussion
 
 
2 hours later…
0celo7
0celo7
2:40 AM
@ACuriousMind Apparently it's a thing to measure mass in pounds
 
user116211
Just giving the definition of the topic doesn't count as a tag wiki, is it so?
 
0celo7
0celo7
3:06 AM
@Slereah Did you know that $S^1\times S^1\times S^1$ does not carry a metric of negative curvature
 
3075
3075
3:20 AM
@BernardMeurer so what finally happened to waterloo?
 
Bernard Meurer
Bernard Meurer
@3075 RIP
 
3075
3075
:(
 
Bernard Meurer
Bernard Meurer
Yeah I was pretty sad
 
3075
3075
what did they say?
 
Bernard Meurer
Bernard Meurer
Like someone sent me an email about a week ago saying they needed an outline of my maths course
I said I didn't have that in hand and:
a. asked what would qualify as that and if a signed letter from my principal would do
b. why was it needed since CS didn't ask nor did any other school
they just replied a couple days ago "Hai no openings, kthxbai"
 
0celo7
0celo7
3:21 AM
RIP me
 
3075
3075
damn.
 
0celo7
0celo7
intro statics problem is kicking my ass
 
dmckee
dmckee
@0celo7 Yes. If you're serious about doing engineering you need to be prepared to deal with pounds-mass versus pounds-weight (and in extreme cases kilograms-mass versus kilograms-weight; argh!).
 
3075
3075
@0celo7 electrostatics?
 
0celo7
0celo7
@3075 no, statics
 
Bernard Meurer
Bernard Meurer
3:22 AM
@3075 Yeah, sigh
 
0celo7
0celo7
did I say electrostatics
 
3075
3075
wow sorry.
i thought you finished first year mechanics.
 
0celo7
0celo7
@dmckee kilogram weight???
@3075 no
 
dmckee
dmckee
@0celo7 I've seen it. But the metric world has moved much farther along the road to making a clear distinction twixt weight and mass.
 
0celo7
0celo7
twixt?
@3075 the picture has an acute angle but I got 270 degrees
 
3075
3075
3:25 AM
nice.
 
dmckee
dmckee
@0celo7 Read you Shakespeare, young man. And get off my lawn.
 
0celo7
0celo7
So I don't know if the drawing is supposed to show me where I'm supposed to measure the angle from
 
dmckee
dmckee
Short for 'betwixt' which means 'between'.
 
0celo7
0celo7
Or if it's actually supposed to be acute
@dmckee Owning land is a crime against humanity, old man
 
Bernard Meurer
Bernard Meurer
@3075 Yeah, but my beer is good though
so it's alright I guess
 
3075
3075
3:27 AM
xD
 
Bernard Meurer
Bernard Meurer
Anyway, I'll catch you guys later, hope you're recovering alright @3075
 
3075
3075
thanks, I'm doing ok.
bye.
 
0celo7
0celo7
who wants to calc a cross product for me
 
user54412
4:27 AM
LIGO just made a sudden announcement to hold a press conference as part of the American Astronomical Society meeting today/tomorrow, 2016-06-15, 10:15 PDT (17:15 UTC). It will be broadcast here.
 
0celo7
0celo7
"JK"
 
user116211
something special?
 
user54412
not sure -- there's excitement in the air
 
user54412
certainly there have been rumors of other signals in the pipeline
 
0celo7
0celo7
ALIENS
 
user116211
4:31 AM
ah!
 
John Rennie
John Rennie
6:04 AM
@ChrisWhite A chat session. We need another special chat session for this. It has to be more mergers. @DavidZ @dmckee can we set up a chat session for this?
 
Secret
Secret
7:04 AM
http://gmwgroup.harvard.edu/research/index.php?page=13

Studying dissipative systems by building them artificially
 
yuggib
yuggib
7:38 AM
here you go...another question on the continuity/differentiability of the wavefunction...
3
Q: Can a physical wavefunction be non-smooth (its first derivative is discontinuous)?

IanDsouzaHere's an argument that might support the statement that such a non-smooth wavefunction is not physical: You cannot add a finite number of smooth functions to get a non-smooth function. By fourier expansion theorem, any function can be expressed as a sum of plane waves (which are smooth with res...

quantum-mechanics fourier-transform
this time I couldn't refrain from answering
how did the ama go yesterday?
(it was too late for me here in Japan)
 
 
1 hour later…
user116211
8:53 AM
@yuggib nice to the extent that you can be our next guinea pig ;P
 
David Z
David Z
@JohnRennie You can set up a chat session, and ask a mod to post an announcement
 
John Rennie
John Rennie
@DavidZ is there any setup needed? I assumed a chat session would just mean us all meeting here at 17:15 UTC.
IIRC that's how it worked for the first LIGO announcement ...
 
user116211
@JohnRennie all in the main page would know about that then if you do that.
 
yuggib
yuggib
@JohnRennie I had a sneak preview of this new announcement... ;-P
 
John Rennie
John Rennie
@yuggib Spill,spill!
If an NDA stops you just tell us if we NEED to watch the live stream!!!
 
user116211
8:57 AM
@yuggib don't make suspense ;P
 
user116211
4 hours ago, by Chris White
certainly there have been rumors of other signals in the pipeline
 
yuggib
yuggib
there was at least another one
of course it was not confirmed, but I have been told to wait for such an announcement
 
John Rennie
John Rennie
Only one :-(
Still that's enough to make the live stream compulsory
 
yuggib
yuggib
well, I know of one
maybe there were more
if they're announcing it probably means they've already another paper accepted
 
John Rennie
John Rennie
@DavidZ: assuming we just meet here at 17:15 UTC, can you make the announcement?
 
yuggib
yuggib
9:02 AM
@MAFIA36790 :-D
 
Feeds
Feeds
Chris White has added an event to this room's schedule.
 
yuggib
yuggib
@MAFIA36790 is there a date already? should I prepare a meta post?
 
John Rennie
John Rennie
@StackExchange Thanks Chris. It never occurred to me to see if I could add the event to the schedule myself.
 
user54412
Hopefully it's something interesting. Given that it's a press conference, and in fact the first press conference since the original, I'm hopeful.
 
yuggib
yuggib
I bet on $\geq 2$ new signals
 
user116211
9:07 AM
@yuggib 1st one: no; 2nd one: really don't know; after seeing last night's AMA, I'm sure we can expect further AMAs but of course, no official announcement either. And moreover, it would be too early, i suppose, to write a meta post. I'm waiting for @vzn for further details on that - when the next AMA would happen or so....
 
user54412
Also, I could have been at that meeting, in San Diego no less. Instead I'm working on a paper. I'm not sure if I'm doing grad school quite right.
 
user116211
:(
 
FrancescoS
FrancescoS
10:15 AM
Is there someone who knows how to use MadGraph to compute partonic cross sections???
Nobody on the internet explains you how to do it without saying a lot of useless details
 
Slereah
Slereah
10:45 AM
Are there any exact solutions to string theory?
Like to the Polyakov action
 
ACuriousMind
ACuriousMind
@Slereah Uh, it's just the string propagating in a straight line.
 
Slereah
Slereah
Hm
What's the gravitational action like in string theory?
 
ACuriousMind
ACuriousMind
The "classical" string is a wholly uninteresting object.
 
Slereah
Slereah
How do I get pants diagrams
I want a cornucopia of pants
 
ACuriousMind
ACuriousMind
@Slereah String theory is an S-matrix theory, not a Lagrangian theory. There is no action.
 
Slereah
Slereah
10:54 AM
Oh
How do I get the gravitational S Matrix, then
 
ACuriousMind
ACuriousMind
You do get various 10D SUGRA theories as the effective QFTs that produce the tree-level ST results, though
@Slereah You "just" insert the appropriate vertex operator into the string perturbation series.
 
Slereah
Slereah
String theory is p. lame
Is there any non-perturbative definition
Also Polchinski defines the string everywhere in his book with a Lagrangian?
 
ACuriousMind
ACuriousMind
@Slereah That's the holy grail - but no one has found it yet. Under "M-theory", "F-theory" and other names, various non-perturbative bits are known, but no one knows what the correct non-perturbative formulation of string theory is.
@Slereah You're mistaking the Polyakov action for "the action" of the theory. The defining object of the dynamics of string theory is the string perturbation series/S-matrx.
 
Slereah
Slereah
Is there a toy model maybe for string theory?
I dunno, 2+1D strings or something
 
ACuriousMind
ACuriousMind
The string theory books are really abysmal at making clear that string theory really is not a QFT
 
Slereah
Slereah
10:58 AM
Or are there no non perturbative definition at all
 
ACuriousMind
ACuriousMind
@Slereah That's called the "non-critical string" and it's harder than the critical one because there you have the Weyl anomaly hence no clue how to properly quantize
 
Slereah
Slereah
Why do people even bother with string theory
 
ACuriousMind
ACuriousMind
@Slereah No, we know various things about the non-perturbative formulation provided it exists, but no one knows its proper definition.
 
Slereah
Slereah
Do we have clues what it is expected to look like?
 
ACuriousMind
ACuriousMind
@Slereah Because it "miraculously" produces many structures that you haven't put in.
@Slereah Yeah, that's what is known as "M-theory"
We know some parts of the spectrum - the BPS objects - for example
But what actual kind of physical theory is might be is probably still open
 
Slereah
Slereah
11:03 AM
Is it ghosts
 
ACuriousMind
ACuriousMind
Is what ghosts?
 
Slereah
Slereah
The real theory
Spooks and haunts
👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻 👻
 
ACuriousMind
ACuriousMind
Well, the "M" might stand for "magic" :P
 
Slereah
Slereah
Man is there any decent quantum gravity theory
That is both well defined and not proven wrong on some level
 
ACuriousMind
ACuriousMind
I think string theory might really be your best bet. It's just very much work in progress, it's as if you had stumbled into QFT before we knew anything about partition functions or path integrals.
That it reproduces QFT at low energy and so generically gives no low energy predictions doesn't help, of course :P
 
Slereah
Slereah
11:09 AM
What's the status of Lorentzian gravity, these days
 
ACuriousMind
ACuriousMind
I have no idea...I have to go to my string theory class, now
 
Slereah
Slereah
Well ask them what the fuck is up
Ask them to make a real theory and stop wasting my time
 
0celo7
0celo7
11:52 AM
good morning plebs
> Haradhan Kumar Mohajan
Assistant Professor, Faculty of Business Studies, Premier University, Chittagong,
Bangladesh
GP paper written by a business prof
OK, seems legit
 
0celo7
0celo7
12:26 PM
I managed to reference my blog in a PSE answer
 
0celo7
0celo7
12:42 PM
@ACuriousMind I told my prof you said a bunch of the stuff I was doing in Milnor and GP can also be done using algebraic topology and he said "where's the fun in that"
 
Secret
Secret
1:30 PM
Interesting, I knew of a martin who also worked with quantum mechanics
@Martin do you by any chance, studied in UNSW in 2014?
 
0celo7
0celo7
::attempts to stalk profs on MO::
 
user116211
1:48 PM
@0celo7 How could it be o.O
 
0celo7
0celo7
I meant GR paper of course
 
user116211
yeah... but still... how could it be o.O
 
user116211
a business prof? hmm...
 
user116211
It's nothing but weird ;\
 
0celo7
0celo7
@JohnRennie How would you answer this?
Please don't take this the wrong way, but I'd like to see a "low tech" argument
 
user116211
1:52 PM
Have anyone saw Feynman's derivation of Lienard-Wiechard Potential?
 
0celo7
0celo7
Never heard of it
 
user116211
It's really giving me a shit.
 
user116211
@0celo7 potentials of a moving charge
 
0celo7
0celo7
yeah, that's not a phrase we use in English
@ACuriousMind Are you the second close vote on this?
 
user116211
He used an elementary idea of discrete summation over a point charge dispersed over a cube moving with a non-relativistic velocity.
 
ACuriousMind
ACuriousMind
1:55 PM
@0celo7 Yes, I don't know what that question is attempting to do.
If he wants to find out whether a metric is flat, why not just compute the curvature?
 
user116211
But actually he overcounts the charge in the volume slices.
 
0celo7
0celo7
@ACuriousMind I'm still not convinced that curvature 0 <==> Euclidean space (locally)
The proof uses the damn black box of black boxes
The Frobenius theorem (PDE version)
 
user116211
There is a PSE post on this... but actually the answer doesn't seem to point the mistake of Feynman.
 
user116211
5
Q: Feynman's proof for Liénard-Wiechert's potential of a moving charge

guillefixFeynman's proof utilizes a geometrical and fundamental integration argument. I like it, except this bit: What makes me unconfortable somehow is that in (c) we are counting in some of the charge we counted at (b). It seems to me that it is this extra counting which makes the potential to be lar...

relativity classical-electrodynamics lienard-wiechert
 
0celo7
0celo7
@ACuriousMind You know a black box is bad when it's used to prove other black boxes!
 
ACuriousMind
ACuriousMind
1:57 PM
::shrugs::
 
0celo7
0celo7
@ACuriousMind what's that shrug for
do you know the proof?
 
ACuriousMind
ACuriousMind
@0celo7 The shrug indicates I don't care
 
0celo7
0celo7
:/
How can you maintain such an attitude
 
Martin
Martin
@Secret: No, in fact, I've sadly never been to Australia, yet.
 
FrancescoS
FrancescoS
Guys, does anyone here knows how to use MadGraph?
 
John Rennie
John Rennie
2:36 PM
@0celo7 What does constant curvature mean in 4D Lorentian manifold. In a 2D Riemannian manifold I'd say it was a constant Gaussian curvature, but I'm unsure how this extends to spacetime.
 
0celo7
0celo7
@JohnRennie In this case it means the curvature scalar is constant
 
John Rennie
John Rennie
@0celo7 The Ricci scalar?
 
0celo7
0celo7
@JohnRennie yes
Same difference
 
John Rennie
John Rennie
Well I suppose I'd compute the Ricci scalar - actually I'd just Google for the result.
 
0celo7
0celo7
@JohnRennie I also believe it follows from Schur's lemma.
you'd probably have to modify it for spacetime
@JohnRennie erm, you don't know the metric yet
this is part of the derivation of it
frick
the idea of sectional curvature doesn't work that well
I know Beem wrote a bunch of papers on it.
because you can have degenerate planes, i.e. planes generated by null vectors
@JohnRennie also do you know what this guy is asking for physics.stackexchange.com/questions/262711/…
 
Obliv
Obliv
2:50 PM
is $\mathcal{O}$ a set?
nvm it is
 
ACuriousMind
ACuriousMind
@Obliv what?
 
0celo7
0celo7
what the hell is that
what kind of question is that
 
Obliv
Obliv
it's an equivalence class
called an orbit i think
 
ACuriousMind
ACuriousMind
Just writing down a letter and asking if it is a set is not a meaningful question.
Never assume your notation is universally known.
 
Obliv
Obliv
@acuriousmind I was going to explain but i realized the answer before i could
 
John Rennie
John Rennie
2:53 PM
@0celo7 I can't see the point of that. Any form of the flat space metric is related to the Minkowksi metric by a coordinate transformation. So the most general form would be written as an arbitrary transformation applied to the Minkwski metric. This strikes me as a boring thing to do.
 
0celo7
0celo7
@JohnRennie *locally minkowski
You could have some weird topology
but yes I agree
not sure what OP wants
 
Obliv
Obliv
Hey lagrange's theorem : "if $G$ is a finite group and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$" only applies when there is a bijective map $\varphi: H \to \mathcal{O}$? Also, I get that this means $\forall h \in H$, $x = hb$ ,for some $x,b\in G$ so that the elements of $H$ divide the elements in $G$. How does this indicate the cardinality of the subgroup divides the cardinality of the group though?
Sorry if i'm not being clear let me know where I don't make sense
 
ACuriousMind
ACuriousMind
3:09 PM
@Obliv You still haven't told us what $\mathcal{O}$ is. Also, Lagrange's theorem always holds.
 
Slereah
Slereah
let's try to learn inverse scattering transform again
I want some of that sine gordon
 
Obliv
Obliv
$\mathcal{O}$ is the equivalence class of an element $x \in G$ under ~ where ~ is defined as $x$~$b$ if $x = hb$ for some $h \in H$ and $x,b \in G$, $H$ is acting on $G$.
 
Slereah
Slereah
Is sine gordon even well defined as a QFT
It has arbitrarily high derivatives
Is it even causal
 
0celo7
0celo7
@Obliv sounds like some coset stuff
 
ACuriousMind
ACuriousMind
@Obliv Not indicating of which element it is such an equivalence class is poor form (that is, at least write $\mathcal{O}_x$). But, apart from that, what exactly is your question?
 
0celo7
0celo7
3:13 PM
@Obliv the best notation is always $[x]$
For instance, I am now classifying the equivalence classes of codimension $p$ Pontryagin manifolds where ~ is framed cobordism
Fun fun
And I'm calling them $[x_i]$
 
Obliv
Obliv
@acuriousmind well suppose you have a bijective map $H \to \mathcal{O}_x$ where $H$ is a subgroup of $G$ acting on $G$. $x \in G$. How can this information alone let us deduce that $|H|$ divides $|G|$? I thought the orbit just shows that the elements of the group being acted on is divisible by the group's elements acting on it.
@0celo7 i have no idea what that means. what's a coset?
 
0celo7
0celo7
@Obliv I don't know either
 
ACuriousMind
ACuriousMind
@Obliv I need not suppose that, the map is simply given by $h\mapsto hx$. This means that all equivalence classes have the same size as $H$. Furthermore all equivalence classes are disjoint (because equivalence classes always are). If $\lvert G \rvert$ is finite, then it's a finite disjoint union (of say $n$ classes), so $\lvert G \rvert = n\lvert H \rvert$.
 
Obliv
Obliv
@acuriousmind $H \to \mathcal{O}_x$ maps elements $h \in H$ to the relation $x$~$h$, correct? So this means $h \to x = hx$ so $h \to hx$ I get that part. Doesn't this just mean the equivalence class of $x \in G$ under ~ is the size of $H$? This is only one element of $G$. Is this supposed to be extended to all elements in $G$?
 
0celo7
0celo7
@Slereah The worst is translating math into GR
 
John Rennie
John Rennie
3:23 PM
@0celo7: maybe this is a silly question, but does the assumption of homogeneity mean there cannot be an inhomogeneous solution? For example the vacuum is homogeneous, and while we do indeed get a homogeneous solution (Minkowski) we also get inhomogeneous solutions (Schwarzschild, Kerr, etc).
 
Slereah
Slereah
especially since GR abuses a lot of math terms
 
ACuriousMind
ACuriousMind
@Obliv What does "maps to the relation $x\sim h$" mean? Your second sentence doesn'T make any sense at all to me. But yes, from the existence of a bijection $H\to \mathcal{O}_x$ we conclude that the equivalence class $\mathcal{O}_x$ has size $\lvert H\rvert$, and since nothing was special about $x$, this holds for all elements in $G$, there is nothing to "extend".
 
Slereah
Slereah
There's like ten homogeneous vacuum solutions
In 4D
In 2D it's like
Flat space
Cylinder
Other cylinder
Torus
Klein bottle
Other Klein bottle
Projective plane
I think, anyway
Might be a little more
 
John Rennie
John Rennie
@Slereah: is that a reponse to my post?
 
Obliv
Obliv
@acuriousmind well the orbit $\mathcal{O}$ is an equivalence class under an equivalence relation. Since the set $\mathcal{O}$ is made of elements that satisfy this relation, I think of it as those elements, which are mapped, having to satisfy this relation. I just wrote it as being mapped to it to save time
 
Slereah
Slereah
3:26 PM
somewhat
 
Obliv
Obliv
Alright thank you :)
 
Slereah
Slereah
I think those are the only homogeneous flat solutions
 
John Rennie
John Rennie
@Slereah I don't see how that answers my question
 
ACuriousMind
ACuriousMind
@Obliv I have no idea what you just said.
 
Slereah
Slereah
I dunno mang
What is your question exactly
I'm not quite sure
 
John Rennie
John Rennie
3:28 PM
@Slereah There are vaccum solutions that don't have constant curvature even though the vacuum is isotropic and homogeneous. Yes?
 
Slereah
Slereah
They have constant curvature
 
Obliv
Obliv
@acuriousmind $\mathcal{O}_x = \{a,b,c,...~|~x$~$a\}$, right? So if $H \to \mathcal{O}_x$ then $h \in H$ must satisfy the condition in the set $\mathcal{O}_x$?
 
Slereah
Slereah
Well, they are Ricci flat, anyway
 
Obliv
Obliv
So I just saved time by saying $h \in H$ was being mapped to $x$~$h$
 
John Rennie
John Rennie
@Slereah The Kretschmann scalar is not constant e.g. in the Schwarzschild metric
 
Slereah
Slereah
3:29 PM
If you mean the Riemann tensor, then yes, you can have vacuum solutions that are not flat
 
Secret
Secret
People often said quantum states are vectors in hilbert space, but why there isn't any states that acts like the zero vector, i.e. when superimposed onto another state, does not change the state?
 
Slereah
Slereah
That's because in >2+1D, the Riemann tensor does not depend exclusively on the Ricci tensor
The 0 vector doesn't have norm 1
Quantum states are rays
0 is not a ray
 
ACuriousMind
ACuriousMind
@Obliv "What?" to the first question. "If $H\to\mathcal{O}_x$" doesn't make any sense,$A\to B$ is not a statement, and no, if $h\sim x$ for some $h\in H$, then $\mathcal{O}_x$ is the equivalence class of the identity.
You have to be way more careful with the way you state things in math.
 
John Rennie
John Rennie
@Slereah So can we have solutions for an isotropic and homogeneous non-vacuum solution that do not have constant curvature? i.e. the various scalars are not constants.
 
Slereah
Slereah
Sure
The De Sitter Schwarzschild solution, for instance
Wait, no, it's not homogeneous
Wait if it's isotropic and homogeneous and non vacuum
It's gonna be either dS or AdS
 
John Rennie
John Rennie
3:32 PM
@Slereah why?
 
Slereah
Slereah
Well you can check it by hand
Apply all the Killing vectors
 
John Rennie
John Rennie
The vacuum can be Minkwski, or Schwarzschild or Kerr or etc etc
 
Slereah
Slereah
You'll find out that the Riemann tensor is entirely determined by a scalar
But Schwarzschild isn't homogeneous
 
John Rennie
John Rennie
Why if you add a small constant energy density is the FLRW metric the only solution?
 
Slereah
Slereah
$r$ isn't a Killing vector at all
 
Obliv
Obliv
3:34 PM
@acuriousmind by definition if the set $H$ is being mapped to a set that imposes a condition, the elements in the set $H$ must satisfy this condition in order for this to be a bijection, right? Also $H \to \mathcal{O}_x$ is supposed to signify mapping of $H$ to $\mathcal{O}$. Lastly, $h$~$x$ must be true in order for the $h$ to be in the orbit of $x$ I thought
 
ACuriousMind
ACuriousMind
@Obliv nononononono. You can map the odd numbers to the even numbers by adding 1. That doesn't mean that odd numbers satisfy the condition of being even.
 
John Rennie
John Rennie
Suppose you make the stress-energy tensor diag(e,0,0,0) for some small constant e. In the limit of e=0 we get all the vacuum solutions. For non-zero but infinitesimally small e the only solution is FLRW. Is that correct?
 
ACuriousMind
ACuriousMind
And I know that $H\to\mathcal{O}$ is supposed to "signify a mapping", it's just that "If $A\to B$" is not a meaningful statement. "If $f:A\to B$ exists" or "If $f:A\to B$ is a bijection" works, but "If $A\to B$" just doesn't mean anything.
 
Slereah
Slereah
Errrr maybe?
 
John Rennie
John Rennie
Insufficient effort?
0
Q: What is the volume of a 4d sphere

DaMegaBoss99What is the volume of a 4d sphere? I've seen so many site that would have answered this question, but all of them have so many numbers that some people (including me) don't understand. So I was hoping somebody here would give me a nice simple explanation.

volume
 
Slereah
Slereah
3:39 PM
Constant dust solution would be FRLW I think yeah
 
ACuriousMind
ACuriousMind
@JohnRennie Not about physics, homework-like question, no effort, take your pick :P
 
Obliv
Obliv
Wait.. so $f: H\to \mathcal{O}_x$ is a bijection isn't implying that $x$~$h$ for all $h \in H$? So there is no significance in $\mathcal{O}_x$'s relation being $x$~$h$ iff $x = hb$ as long as it's an equivalence relation and this map $f$ is a bijection, $|H|$ divides $|G|$? I feel stupid now @acuriousmind
 
ACuriousMind
ACuriousMind
@Obliv I'll state this once again properly: The bijection is given by $f : H\to\mathcal{O}_x, h \mapsto f(h)=hx$ (prove that this is a bijection!). This means that $\lvert \mathcal{O}_x\rvert = \lvert H \rvert$ for all $x\in G$. Since $G$ is the disjoint union of $n$ such equivalence classes, $\lvert G\rvert = n\lvert H \rvert$.
Don't try anything else until you have understood this argument.
 
John Rennie
John Rennie
@Slereah well yes. When I started thinking about it maybe was the best I could come up with as well.
 
GPhys
GPhys
so a little bird told me I should watch the new gravity wave conference today
 
John Rennie
John Rennie
3:45 PM
@GPhys Absolutely, more merger detections!! (Allegedly :-)
Though it doesn't start for 90 minutes yet.
 
Slereah
Slereah
Pretty sure that all tensor fields related to the metric are identical for a homogeneous and isotropic spacetime, though
 
GPhys
GPhys
fun
 
John Rennie
John Rennie
@Slereah and yet the vacuum is homogeneous and isotropic ...
 
Obliv
Obliv
What I'm about to say might sound really stupid and you might get mad at me @acuriousmind but I must say it. It's a bijection because by definition $f(h) = (x,h) \in \mathcal{O}_x = x$~$h = hx$ and all of the domain $H$ is being mapped to the equivalence class so all $h\in H$ map to $hx$. What's wrong with this reasoning?
 
Slereah
Slereah
@JohnRennie No
There are plenty of vacuum solutions which are not
 
John Rennie
John Rennie
3:51 PM
You misunderstand me. In a vaccum the stress-energy tensor is isotropic and homogeneous, and in an FLRW spacetime the stress-energy tensor is isotropic and homogeneous. So why are there many vacuum solutions and only one FLRW metric?
 
Slereah
Slereah
Oh, you mean the stress energy tensor
Hm
Also when you say "isotropic and homogeneous", you mean just the spatial part, right?
 
John Rennie
John Rennie
@Slereah yes
 
Slereah
Slereah
I'd say there are more than one solution type for the same homogeneous and isotropic stress energy tensor
Symmetries of the SET are symmetries of the Ricci tensor
Those will only determine the Ricci tensor
The Weyl tensor is still free
 
KISHORE_ZE
KISHORE_ZE
hello
 
John Rennie
John Rennie
@KISHORE_ZE Hi. Come for the gravity wave chat?
 
KISHORE_ZE
KISHORE_ZE
3:56 PM
I suppose
LIGO Gravity WAve lab press conf.
 
ACuriousMind
ACuriousMind
@Obliv I have no idea what $f(h)=(x,h)\in\mathcal{O}_x = x\sim h = hx$ is supposed to mean, and I don't understand your reasoning at all. What is $(x,h)$?! Why is it $f(h)$, or an element of $\mathcal{O}_x$.
 
KISHORE_ZE
KISHORE_ZE
Wave*
I'm new to stack physics
why dosen't the formatting show up for me?
 
0celo7
0celo7
@JohnRennie the assumption of homogeneity is cosmology
Schwarzschild is not a cosmological model
 
John Rennie
John Rennie
@KISHORE_ZE Mathjax isn't supported in the chat by default.
 
0celo7
0celo7
@ACuriousMind damn you're pretty harsh
 
KISHORE_ZE
KISHORE_ZE
3:58 PM
Oh
Thanks
 
John Rennie
John Rennie
 
0celo7
0celo7
@JohnRennie homogeneous basically means that you can find coordinates in which the metric components are constant
@JohnDuffield still does not understand this
 
Slereah
Slereah
He's not talking about symmetries of the metric tho
Only of the SET
 
John Rennie
John Rennie
@Slereah Yes!!! (SET = stress-energy tensor)
 
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